aramutselv

2022-01-06

$\sqrt[31]{12}+\sqrt[12]{31}$ is irrational?

Elois Puryear

I would assume that $\sqrt[31]{12}+\sqrt[12]{31}$ is rational and try to find a contradiction
However, I don't know where to start. Can someone give me a tip on how to approach this problem?

Jillian Edgerton

Let $\mathbb{Q}\left(\alpha \right)$ denote the smallest field containing $\mathbb{Q}$ and $\alpha$
The theory of field extensions tells us that $\mathbb{Q}\left(\sqrt[31]{12}\right)$ has degree $31$ over $\mathbb{Q}$, $\mathbb{Q}\left(\sqrt[12]{31}\right)$ has degree 12 over $\mathbb{Q}$ , and, because $\left(31,12\right)=1$, we have
$\mathbb{Q}\left(\sqrt[31]{12}\right)\cap \mathbb{Q}\left(\sqrt[12]{31}\right)=\mathbb{Q}$
If $\sqrt[31]{12}+\sqrt[12]{31}$ were a rational number, we would have
$\sqrt[31]{12}\in \mathbb{Q}\left(\sqrt[31]{12}\right)\cap \mathbb{Q}\left(\sqrt[12]{31}\right)=\mathbb{Q}$ But $\sqrt[31]{12}$ is not rational, contradiction.

Vasquez

It is known that algebraic integers are closed under addition, subtraction, product and taking roots.
Since 12 and 31 are algebraic integers, so does their roots $\sqrt[31]{12},\sqrt[12]{31}$. Being the sum of two such roots, $\sqrt[31]{12}+\sqrt[12]{31}$ is an algebraic integer.
It is also known that if an algebraic integer is a rational number, it will be an ordinary integer. Notice
$2<\sqrt[31]{12}+\sqrt[12]{31}<\sqrt[31]{{2}^{4}}+\sqrt[12]{{2}^{5}}={2}^{\frac{4}{31}}+{2}^{\frac{5}{12}}<2\sqrt{2}<3$
$\sqrt[31]{12}+\sqrt[12]{31}$ isn't an integer and hence is an irrational number.

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